prove a function is an interior point of a subset of the metric space $C[-1,1]$ Let $Z$ be the subset of the metric space $C[−1, 1]$ consisting of functions that are equal to zero somewhere: $Z = \{f \in C[-1,1]: f(z) = 0$ for at least one $z \in [-1,1]\} $. Let $g \in Z $ with $g(0) =0$. Show that, if $g$ is differentiable at $0$ and $g'(0) \neq 0 $, then $g$ is in fact an interior point of $Z$. The metric space $C[-1,1]$ equipped with metric d: $d(f,g) = \sup\{\vert f(x)-g(x) \vert: x \in [-1,1]|\} $.
My attempt: 
I find out Z is not open, since functions without $f(z) = 0$ could also in the ball of $f$. Then i have no idea how to start with the derivative.
 A: Since $g'(0)\neq 0$ it must be say positive. Then any $f$ sufficiently close to $g$ in your metric will have some positive and some negative value near $0,$ so using IVT will be $0$ near $0.$ (And thus $f$ will be in the set $Z.$)
Note that the above (once one specifies say $g'(0)>0$) is using that $g$ itself is positive for $x \in (0,\delta)$ and negative in $(-\delta,0)$ for some sufficiently small positive $\delta.$
[If anyone read this before, I had switched $f,g$ but now have put them back as they are in the post, which didn't mention $f,$ I'm just using $f$ as a function sufficiently close to $g$ in the sup metric.]
A: As $g$ is differentiable at $0$ with $g^\prime(0) = c \neq 0$, for all $\epsilon >0$, it exists $\delta > 0$ such that for $\vert x \vert \le \delta$, you have
$$\vert g(x) - c x \vert < \epsilon \vert x \vert.$$ Even if it means changing $g$ into $-g$, we can suppose $c > 0$. Let's take $\epsilon = \frac{c}{3}$. One can find $\delta_0 >0$ such that for $\vert x \vert \le \delta_0$ you have
$$\vert g(x) - c x \vert < \frac{c}{3} \vert x \vert.$$ That implies
$$g(\delta_0) > \frac{2}{3} c \delta_0 \text{ and } g(-\delta_0) <- \frac{2}{3} c \delta_0$$
Now for $d(f,g) < \frac{1}{3} c \delta_0$, you'll have
$$f(\delta_0) > \frac{1}{3} c \delta_0 \text{ and } f(-\delta_0) <- \frac{1}{3} c \delta_0$$ For $f \in C[−1, 1]$, $f$ will vanish in the interval $(-\delta_0,\delta_0)$ which proves that $f \in Z$. Therefore the open ball centered on $g$ and of radius $\frac{1}{3} c \delta_0$ is included in $Z$. Proving the expected result.
